The Presentation of The Yamabe
Problem
Mingfeng Zhao
April 27, 2011
Definition 1 Let M be a smooth manifold, we
say two smooth Riemannian metrics g and h on
M are conformally equivalent if g = λh for some
λ ∈ C ∞(M ) and λ > 0.
Denote the conformal class of the metric g by
[g] = {g̃ : g̃ is conformal to g}
Definition 2 Let (M, g) and (N, h) be two Riemannian manifolds, we say a smooth mapping
f : M → N is a conformal mapping between M
and N if
1. F is a diffeomorphism;
2. F ∗h = λg for some λ ∈ C ∞(M ) and λ > 0.
In this case, we say (M, g) and (N, h) are conformal
equivalent.
1
1960, Hidehiko Yamabe wanted to solve the following question:
For n ≥ 2, (M, g) is any compact smooth ndimensional Riemannian manifold without
boundary, is there a conformal metric g̃ of
g such that (M, g̃) has a constant scalar curvature?
The answer to this question is YES, but Yamabe’s proof is not correct or incomplete, the error of Yamabe’s proof was pointed out by Neil S.
Trudinger in 1968.
2
For n = 2, we have the following uniformization
theorem in complex analysis, which was conjectured by Felix Klein in 1883, and was proved rigorously by Henri Poincare in 1907:
Theorem 1 Let (M, g) be a smooth compact 2dimensional Riemannian manifold. Then there is a
metric g̃ which is conformal to g such that (M, g̃)
has constant Gaussian curvature.
Notice
the Scalar curvature = 2 × the Gaussian curvature
3
Let (M, g) be n-dimensional Riemannian manifold,
and
• G = det (gij )
• D = the Levi-Civita connection
• Dk = the k-times covariant derivative operator
• ∇ = the gradient operator
• Γkij = the Christoffel symbols
• div = the divergent operator
• ∆g = the Laplace-Beltrami operator
• R = the scalar curvature
4
If we set g̃ = e2ug, then g̃ij = e2ugij , and g̃ ij =
e−2ug ij . By a long computation, we can discover
R̃jk
h
i
2
= Rjk − ∆g u + (n − 2)|∇u| gjk
+(n − 2)
∂u ∂u
+ (2 − n)uj;k
j
k
∂x ∂x
where D2u = uj,k dxj ⊗ dxk , and
∂ 2u
∂u
i
uj;k =
−
·
Γ
jk
∂xj ∂xk ∂xi
By contracting with g̃ = e2ug, obtain
n−2
2u
|∇u|2
e R̃ = R + 2(1 − n) ∆g u +
2
Where
1 ∂ √ ij ∂u
∆g u = √
Gg
i
∂xj
G ∂x
√ ij
2
∂ u
1 ∂( Gg ) ∂u
= g ij i j + √ ·
·
i
∂x ∂x
∂x
∂xj
G
5
When n = 2, we get
−2∆g u + R = e2uR̃
When n ≥ 3, then n−2
2 > 0, and for any λ 6= 0, the
following identity holds:
1 −λu
2
∆g eλu
∆g u + λ|∇u| = e
λ
Here we use λ = n−2
2 , then
n−2 u
4(n − 1) − n−2 u
2u
2
e R̃ = R −
∆g e 2
e
n−2
Let v =
n−2 u
e 2 ,
4
2 ln v, and e2u = v n−2 ,
that is, u = n−2
then we can get g̃ =
4
v n−2 g,
and
n+2
4(n − 1)
−
∆g v + Rv = R̃v n−2
n−2
6
In a summary, if n = 2, and the conformal metric
g̃ = eug for some u ∈ C ∞(M ), then
−∆g u + R = R̃eu
If n ≥ 3, and the conformal metric g̃ =
some u ∈ C ∞(M ) and u > 0, then
(1)
4
n−2
g
u
n+2
4(n − 1)
−
∆g u + Ru = R̃u n−2
n−2
for
(2)
So the Yamabe’s question is equivalent to solve
the above two semilinear elliptic PDEs such that R̃
is a constant, that is, for some given constant R̃,
can we find a positive, smooth solution of (2) or
smooth solution of (1)?
7
Look at the case n = 2, if K denotes the Gaussian
curvature, we know
R = 2K
By integrating the PDEs (1), we can get
−
Z
M
∆g u dVg +
Z
M
R dVg =
Z
M
R̃eu dVg
By Stokes’ Theorem, and ∂M = φ, we know
Z
M
∆g u dVg =
Z
M
div(∇u) dVg =
Z
∂M
g(∇u, N ) dVg = 0.
By Gauss-Bonnet Theorem, we know
Z
M
R dVg = 2
Z
M
KdVg = 4πχ(M )
So we have
Z
M
R̃eu dVg = 4πχ(M )
8
To solve the above PDEs, Yamabe introduced the
following functional (called Yamabe functional):
Let M be a compact smooth manifold, denote
M = {all Riemannian metrics on M }
Then we define E : M → R as for any g ∈ M, then
R
Rg dVg
E(g) = RM
n−2
( M dVg ) n
When n = 2, by Gauss-Bonnet Theorem, obtain
E(g) = 4πχ(M ).
The Yamabe functional is the normalized EinsteinHilbert functional or total scalar curvature functional: H : M → R given by for any g ∈ M, then
H(g) =
Z
M
Rg dVg
Easy to show that (M, g) is Einstein iff g is a critical
point of H.
9
In the following, without special words, we always
assume n ≥ 3.
For Riemannian manifold (M, g), we consider the
restriction Qg of E on the conformal class of g: so
4
for any u ∈ C ∞(M ), and u > 0, and g̃ = u n−2 g, then
we define
R
R̃ dVg̃
Qg (u) = E (g̃ ) = M
n−2
R
n
dV
g̃
M
By the PDE (2) and the Stokes’s Theorem, we can
get
i
R h 4(n−1)
2
2
dVg
M
n−2 |∇u| + Ru
Qg (u) =
n−2
2n
R
n
n−2 dVg
u
M
(3)
10
Notice that the critical point u of Qg satisfies the
Euler-Lagrange equation:
4(n − 1)
−
∆g u + Ru =
n−2
i
R h 4(n−1)
2
2
dVg
M
n−2 |∇u| + Ru
n−2
kuk 2n2n
L n−2 (M )
Set
R̃ =
i
R h 4(n−1)
2
2
dVg
M
n−2 |∇u| + Ru
n−2
kuk 2n2n
L n−2 (M )
.
Hence the critical point of Qg gives our result.
11
·
n+2
u n−2
n > 1, and 1 + 1 = 1, so by Holder’s
Since n−2
n
n
2
n−1
inequality, we know
Z
2
Ru dVg M
≤ kRkL∞(M )
Z
M
u2 dVg
≤ kRkL∞(M ) (Vol
= Cg
Z
M
2n
u n−2
2
M )n
Z
M
n
2· n−2
u
n−2
dVg
n−2
dVg
n
.
So Qg has lower bound, then define
Y (M, g) = inf E(g̃) =
g̃∈[g]
inf
u∈C ∞ (M ),u>0
Qg (u)
We call Y (M, g) the Yamabe (conformal) invariant
of the Riemannian manifold (M, g).
By the definition of Y (M, g), if g̃ is conformal to g,
then
Y (M, g) = Y (M, g̃).
12
n
For any k ≥ 1, and 1 ≤ p < ∞, for any u ∈ C ∞(M ),
we can define the following norm:

kukW k,p = 
k Z
X
i=0 M
1
p
|Diu|p dVg 
where
|Diu|2 = g(Diu, Diu).
The Sobolev spaces W k,p(M ) is the completion of
C ∞(M ) under the the above norm.
For k ≥ 0, 0 < α ≤ 1, the Holder space C k,α(M ) is
the set of all functions u ∈ C k (M ) such that
|Dk u(x) − Dk u(y)|
sup
< ∞.
α
|x − y|
x,y∈M,x6=y
Notice that the Sobolev space W k,p(M ) and the
Holder space C k,α(M ) are Banach spaces.
13
Theorem 2 (Sobolev Embedding Theorem)
Let k ≥ 1 be integer, and 1 ≤ p < ∞. If kp < n,
then
np
W k,p(M ) ,→ L n−kp (M ).
np
Moreover, if 1 ≤ q < n−kp
, then
W k,p(M ) ,→ Lq (M ) is compact.
np
But W k,p(M ) ,→ L n−kp (M ) is not compact.
If kp > n, then
W k,p(M ) ,→ C 0,α(M ) for some 0 < α ≤ 1.
14
By the Sobolev embedding theorem, we can show
Qg (u) is continuous in W 1,2(M ).
Since Qg (u) = Qg (|u|) for all u ∈ C ∞(M ), and any
non-negative function in W 1,2(M ) can be approximated by positive function in W 1,2(M ). So we
have
Y (M, g) =
inf
u∈C ∞ (M )
Qg (u)
15
In the representation of Qg , the Sobolev critical
2n comes out, this means,
exponent n−2
2n
W 1,2(M ) ,→ L n−2 (M )
is not a compact embedding. So we can not directly use variational method to find minimizer of
Qg .
Hence Yamabe introduced the following subcritical
2n , consider
functional: For any 2 ≤ p ≤ n−2
4(n−1)
2 + Ru2
|∇u|
M
n−2
Qg;p(u) =
2
R
p
( M u dVg ) p
R
h
i
dVg
(4)
Since 2 ≤ p, by Holder’s inequality, Qg has lower
bound, then define
Y (M, g; p) =
inf
u∈C ∞ (M )
Qg;p(u)
In particular,
Qg;
2n
n−2
= Qg ,
and
Y
2n
= Y (M, g).
M, g;
n−2
16
Y (M, g; p) has the following properties:
Theorem 3 (Aubin, 1976) If the smooth compact Riemannian manifold (M, g) such that
R
M dVg = 1, hthen |Y
i (M, g; p)| is non-increasing
2n .
function of p ∈ 2, n−2
Moreover, if Y (M, g) ≥ 0, then Y (M, g; p) is left
continuous.
R
Since M
Proof of Theorem:
dVg = 1, then
kukLp(M ) ≤ kukLq (M ) if p ≤ q. And
Qg;q =
kuk2
Lp (M )
kuk2
Lq (M )
· Qg;p
17
Notice that the critical point up of Qg;p satisfies the
Euler-Lagrange equation:
4(n − 1)
∆g up + Rup = R̃upp−1
−
n−2
(5)
Theorem 4 (Yamabe, 1960) For any 2 ≤ p <
2n , there exists a positive, smooth function u
p
n−2
solves the PDE (5), and such that
Qg;p(up) = Y (M, g; p),
and
kupkLp(M ) = 1.
Proof of Theorem : Taking the minizing sequence
∞ (M )
{uk;p}∞
⊂
C
k=1
such that
kuk;pkLp(M ) = 1,
1,2 (M ).
which implies {uk;p}∞
k=1 is bounded in W
Then there exists a subsequence is weak convergent. By compact embedding, get the limiting
function up.
Finally, by the standard bootstrap method to show
up ∈ C 2,α(M ). Then by using the strong maximum
principle, we conclude u > 0. Again by the bootstrap method to show u ∈ C ∞(M ).
18
Yamabe stated but not prove that up’s are uni2n . But in 1968 Trudinger
form bounded, as p % n−2
pointed out that this is not true in general. And
Trudinger can prove:
Theorem 5 (Trudinger, 1968; Aubin, 1976)
Let g be the standard metric on S n, that is, induced by the embedding i : S n ,→ Rn+1. Suppose
2n }
Y (M, g) < Y (S n, g), and let {up : 2 ≤ p < n−2
be the solutions in Theorem 4. Then there exists
2n , and q > 2n , and C > 0
constants 2 ≤ p0 < n−2
0
n−2
such that
kupkLq0 (M ) ≤ C
2n .
whenever p0 ≤ p < n−2
19
By the bootstrap method, the Arzela-Ascoli Theorem, we can show a subsequence up converges to
some u is the C 2-norm. Also by the nonincreasing
and left continuity of |Y (M, g; p)|, we can conclude
that
2n } be the soluTheorem 6 let {up : 2 ≤ p < n−2
2n , there exists
tions in Theorem 4. Then as p % n−2
a subsequence converges uniformly to a positive,
smooth function u such that Qg (u) = Y (M, g) and
n+2
4(n − 1)
−
∆g u + Ru = Y (M, g)u n−2
n−2
In particular, the metric g̃ =
scalar curvature Y (M, g).
4
u n−2 g
has constant
20
By the ”renormalization“ method, Uhlenbeck can
show
Theorem 7 There exists a positive, smooth function u on S n satisfying Qg (S n) = Y (S n, g).
Combine with Obata’s Theorem:
Theorem 8 (Obata, 1971) Let g̃ be a conformal
metric to the standard metric g on S n. Then there
is a conformal mapping Φ : (S n, g̃) → (S n, g) such
that
Φ ∗ g = λg̃
for some positive constant λ > 0.
Proof of Theorem: First show (S n, g̃) is Einstein,
that is, the traceless Ricci tensor S = Ric − R
n g ≡ 0.
Then show the Weyl tensor W ≡ 0.
21
Hence, we get the Yamabe solution on S n:
Theorem 9 The Yamabe function Qg on (S n, g)
is minimized by constant multiples of the standard
metric g and its images under conformal diffeomorphisms.
Moreover, these are only metric conformal to the
standard one on S n that have constant scalar curvature.
Since the scalar curvature R on S n is equal to n(n−
1), then
Y (S n, g) = Qg (1)
R
n n(n − 1) dVg
= S
n−2
R
n
dV
n
g
S
2
= n(n − 1) (Vol S n) n
22
For the remaining case, we have
Theorem 10 (Aubin, 1976) If n ≥ 6, and (M, g)
is not locally conformally flat, then Y (M, g) <
Y (S n, g).
In particular, (M, g) can be conformally deformed
to constant scalar curvature .
Theorem 11 (Schoen, 1984) If n = 3, 4 or 5,
of if (M, g) is locally conformally flat, then either Y (M, g) < Y (S n, g), or (M, g) is conformal to
(S n, g).
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Theorem 12 (Aubin, 1976) If (M, g) is any compact Riemannian manifold of n ≥ 3, then Y (M, g) ≤
Y (S n, g).
Proof of Theorem: Use the function
uα(x) =
|x|2 + α2
! 2−n
2
α
x ∈ Rn
to construct a function u on M such that
Qg (u) ≤ (1 + C)(Y (S n, g) + Cα).
Taking α → 0, and → 0, we get
Y (M, g) ≤ Y (S n, g).
Notice that
k∇uαk2
L2 (Rn )
kuαk2
2n
n−2
L
( Rn )
n−2
=
· Y (S n, g)
2(n − 1)
24
Consider the Stereographic projection
σ : S n\{(0, · · · , 0, 1)} −→ Rn,
By computation, we have
4
n−2
(σ −1)∗g = 4u1
(x)dxi ⊗ dxi.
So σ is a conformal mapping.
Consider the dilation δα : Rn → Rn, then
4
n−2
∗ (σ −1 )∗ g = 4u
δα
α
dxi ⊗ dxi
For any φ ∈ C ∞(M ), we define
φ = u1(σ −1)∗φ
So we can get
4
4
g
(σ −1)∗(φ n−2 ) = 4φ n−2 dxi ⊗ dxi.
25
By the conformal invariance of Qg , we can conclude
4(n−1)
2
n n−2 |∇φ| dx
R
Y (S n, g) =
inf
n−2
φ∈C ∞ (S n ) R
2n
n
n−2
|φ|
dx
Rn
R
Since C0∞(Rn) is dense in C ∞(Rn) in the W 1,2(Rn)norm, then we get
R
Y (S n, g) =
inf
∞
Rn
4(n−1)
2 dx
|∇φ|
n−2
n−2
2n
φ∈C0 (S n ) R
n−2 dx
Rn |φ|
n
So Y (S n, g) gives the best Sobolev constant.
26